then :
A) P is independent of x
B) \[P=(x-a)\,\,(x-b)\,\,(x-c)\]
C) \[P=abc\,\,{{x}^{3}}\]
D) none of these
Correct Answer: A
Solution :
\[{{c}_{1}}\to {{c}_{1}}-{{c}_{2}}\] and \[{{c}_{2}}\to {{c}_{2}}-{{c}_{3}}\] \[\Rightarrow \] two identical columns \[\Rightarrow \] A
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